# Computing the equivariant cohomology of a specific $(\mathbb{Z}/2\mathbb{Z})^2$-space

In the paper On the Castelnuovo-Mumford regularity of the cohomology ring of a group, Symonds describes the following space.

Let $G = (\mathbb{Z}/2\mathbb{Z})^2 = \{1,a,b,ab\}$ be an elementary abelian $2$-group, and let $Z = \{ z_1, \dots, z_6\}$ be a discrete space with six elements. Let $G$ act on $Z$ such that each rank one subgroup of $G$ is a stabilizer of some element of $Z$. If I'm not mistaken, this means that the action is for example given as follows: $$a \cdot z_1 = b \cdot z_1 = z_2;\\ a \cdot z_3 = z_4; \; b \cdot z_3 = z_3; \; b \cdot z_4 = z_4;\\ b \cdot z_5 = z_6; \; a \cdot z_5 = z_5; \; a \cdot z_6 = z_6.$$

Let $SZ$ be the suspension of $Z$, with the induced $G$-action.

Question. How to compute the equivariant cohomology $H^*_G(SZ)$ as an $H^*_G$-module (with coefficients $\mathbb{F}_2$)?

I believe I could compute the equivariant cohomology $H^*_G(Z)$, using the Mayer–Vietoris sequence for equivariant cohomology (it split as a disjoint union of three $G$-spaces), the Künneth theorem, and the known results about the equivariant cohomology of spaces with a free (resp. trivial) action. But unfortunately there's no suspension isomorphism for equivariant cohomology, as far as I can tell...

Of course I know the definition of equivariant cohomology; the classifying space $BG$ is $(\mathbb{RP}^\infty)^2$, and $H^*_G = \mathbb{F}_2[x,y]$ is a polynomial algebra on two variables of degree $1$. The total space of the universal bundle is $(S^\infty)^2$ with $a$ (resp. $b$) acting by the antipodal action on the first (resp. second) factor, and $H^*_G(SZ) = H^*(EG \times_G SZ)$. But this isn't exactly helpful, or at least I don't see how to compute the equivariant cohomology just from that.

• Maybe it's better ask on MathOverflow. – Vincenzo Zaccaro Apr 11 '17 at 9:48

The standard way to compute equivariant cohomology of a $G$-space $X$ is to use the spectral sequence of the fibration $$X\to EG\times_G X\to BG,$$ where the projection is induced by $X\to \ast$. With $\mathbb{F}_2$ coefficients this takes the form $$E_2^{p,q} = H^p(BG; H^q(X;\mathbb{F}_2))\Rightarrow H^*_G(X;\mathbb{F}_2).$$ In your case, the space $X=SZ$ is connected and one-dimensional, so there are only two non-trivial rows: $q=0$ (which just gives $H^*(BG;\mathbb{F}_2) = H^*_G$) and $q=1$ (the cohomology of $G=\mathbb{Z}/2\times\mathbb{Z}/2$ with coefficients in the module $H^1(SZ;\mathbb{F}_2)$). Hence the $E_2$-term is computable.
There is only one possible non-trivial differential $$d_2:H^p(BG;H^1(SZ;\mathbb{F}_2))\to H^{p+2}(BG;H^0(SZ;\mathbb{F}_2))=H^{p+2}(BG;\mathbb{F}_2).$$
However, the fact that $SZ$ has $G$-fixed points (the suspension points) implies that the above fibration admits a section, and so the induced map $H^*(BG;\mathbb{F}_2)\to H^*_G(SZ;\mathbb{F}_2)$ is injective. This agrees with the edge homomorphism of the spectral sequence, implying that $d_2$ is in fact trivial. Hence $E_2=E_\infty$. There are no extension problems, since we are using field coefficients, so this gives the equivairiant cohomology as a $\mathbb{F}_2$-module.
I believe it also gives that $H^*_G(SZ;\mathbb{F}_2)$ is free as an $H^*_G$-module.
• @NajibIdrissi: You're welcome. I should add that I've had a change of mind regarding the last sentence of my answer. It would be true (that $H^*_G(SZ;\mathbb{F}_2)$ is free as a $H^*_G$-module) if $G$ acts trivially on $H^1(SZ;\mathbb{F}_2)$. – Mark Grant Apr 16 '17 at 11:54